Question Number 212557 by tri26112004 last updated on 17/Oct/24
Answered by Ghisom last updated on 17/Oct/24
![cos x =c sin x =(√(1−c^2 )) ((√(1−c^2 )))^8 +c^8 = 2c^8 −4c^6 +6c^4 −4c^2 +1= =2(c^4 −c^2 +1+((√2)/2))(c^4 −c^2 +1−((√2)/2))= [c^4 −c^2 =−c^2 (1−c^2 )=−cos^2 x sin^2 x =((cos 4x −1)/8)] =2(((cos 4x)/8)+(7/8)+((√2)/2))(((cos 4x)/8)+(7/8)−((√2)/2))= =((cos^2 4x)/(32))+((7cos 4x)/(16))+((17)/(32))= =((1+cos 8x)/(64))+((7cos 4x)/(16))+((17)/(32))= =((cos 8x)/(64))+((7cos 4x)/(16))+((35)/(64))](https://www.tinkutara.com/question/Q212571.png)
$$\mathrm{cos}\:{x}\:={c} \\ $$$$\mathrm{sin}\:{x}\:=\sqrt{\mathrm{1}−{c}^{\mathrm{2}} } \\ $$$$\left(\sqrt{\mathrm{1}−{c}^{\mathrm{2}} }\right)^{\mathrm{8}} +{c}^{\mathrm{8}} = \\ $$$$\mathrm{2}{c}^{\mathrm{8}} −\mathrm{4}{c}^{\mathrm{6}} +\mathrm{6}{c}^{\mathrm{4}} −\mathrm{4}{c}^{\mathrm{2}} +\mathrm{1}= \\ $$$$=\mathrm{2}\left({c}^{\mathrm{4}} −{c}^{\mathrm{2}} +\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\left({c}^{\mathrm{4}} −{c}^{\mathrm{2}} +\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)= \\ $$$$\:\:\:\:\:\left[{c}^{\mathrm{4}} −{c}^{\mathrm{2}} =−{c}^{\mathrm{2}} \left(\mathrm{1}−{c}^{\mathrm{2}} \right)=−\mathrm{cos}^{\mathrm{2}} \:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:=\frac{\mathrm{cos}\:\mathrm{4}{x}\:−\mathrm{1}}{\mathrm{8}}\right] \\ $$$$=\mathrm{2}\left(\frac{\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{8}}+\frac{\mathrm{7}}{\mathrm{8}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\left(\frac{\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{8}}+\frac{\mathrm{7}}{\mathrm{8}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)= \\ $$$$=\frac{\mathrm{cos}^{\mathrm{2}} \:\mathrm{4}{x}}{\mathrm{32}}+\frac{\mathrm{7cos}\:\mathrm{4}{x}}{\mathrm{16}}+\frac{\mathrm{17}}{\mathrm{32}}= \\ $$$$=\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{8}{x}}{\mathrm{64}}+\frac{\mathrm{7cos}\:\mathrm{4}{x}}{\mathrm{16}}+\frac{\mathrm{17}}{\mathrm{32}}= \\ $$$$=\frac{\mathrm{cos}\:\mathrm{8}{x}}{\mathrm{64}}+\frac{\mathrm{7cos}\:\mathrm{4}{x}}{\mathrm{16}}+\frac{\mathrm{35}}{\mathrm{64}} \\ $$